Dumbbell maths

After a lifetime of mathematics it is always interesting to come across a new slant on a mathematical topic. This insight about dumbbell mathematics comes from the 1988 mathematical exploration by David Wells called Hidden Connections, Double Meanings.

Much mathematics arises out of playing. Let us take an arbitrary quadrilateral, mark the midpoints of two opposite sides, join these midpoints and take the midpoint of the resulting segment to obtain point X. What can we say about this point? If the quadrilateral had been a square or a rectangle, X would be the centre.

We could equally well start with the other pair of sides to obtain a point Y. Are the points X and Y the same? They certainly appear to be. Could we show this?

We use an argument from physics. Our picture below shows a line segment and a dumbbell with two 1 kg weights connected by a very light (‘weightless’) rod. This rod has a midpoint, and we could balance the dumbbell at this midpoint. In fact, the weight acting at this midpoint would be 2 kg.

Now let us carry out a similar procedure with the quadrilateral. We place a 1 kg weight at each vertex, replace the edges by thin weightless rods, and in fact cover the quadrilateral with a thin rigid weightless surface, as we will finally want to balance the quadrilateral at an interior point.

Now where will the balance point of this quadrilateral be? Our intuitive understanding tells us that wherever it is, the balancing point will be unique. We start by taking an opposite pair of sides. These form two dumbbells (below): the left hand dumbbell being equivalent to a 2 kg weight at X, the right being equivalent to a 2 kg weight at Y. We now have a new dumbbell XY with midpoint T. This will balance at the point T with a weight of 4 kg.

We can similarly repeat the dumbbell argument using the other pair of opposite sides, to obtain a balance point for the quadrilateral, presumably at point

Actually, the dumbbell argument does not add significantly to our original midpoint argument, except that we may well now be convinced from our physical understanding that the balance point of the quadrilateral is unique. Assuming it has a balance point, how we determine that point is immaterial.

So it is now with some conviction that we assert that:

For any quadrilateral, the segments joining the midpoints of opposite sides, themselves have a common midpoint: the centre of the quadrilateral.

But there is more!

Tetrahedra

Because we have started with a planar quadrilateral, we have assumed that our four weights have all been in the same plane. However, there is nothing in our argument which uses this assumption.

Suppose the four weights are not coplanar. They now lie at the vertices of a tetrahedron in space. Suppose we repeat our previous argument to find the centre. The four drawn segments give us two opposite pairs of sides as before, but now we notice something else: the tetrahedron actually has six sides forming three opposite pairs! The argument carries through in the same way, as does the physical realisation of a unique balance point.

This gives:

For any tetrahedron, the three segments joining the midpoints of opposite edges, themselves have a common midpoint: the centre of the tetrahedron.

You might like to verify that we actually obtain the centroid here (you can use a simple vector argument). But, there is even more!

Quadrilateral–again

Back in the plane, we now realise that the four vertices of a quadrilateral determine not just four lines (sides), but six lines: the four sides plus the two diagonals. Look at the diagram below:

It is the same as the tetrahedral diagram, but we now visualise it as lying in the plane. Our work with the tetrahedron shows that there is another relationship for the quadrilateral in the plane which we have missed. We state it now:

For any quadrilateral, the midpoint of the segment joining the midpoints of the diagonals, lies at the centre of the quadrilateral.